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Unformatted text preview: Copyright UC Regents Davis campus, 200511. All Rights Reserved. May not be redistributed without prior written consent of course instructor. Solutions to Assignment 2 Note: This problem is similar to Problem 120 on Page 28 of your Segel book. Problem 25 The ionization of H 3 PO 4 is shown below: H + + H 2 PO 4 H + + HPO 4 = H + + PO 4 H 3 PO 4 Part a) In this part of the problem you will use the K a equation as follows to determine the hydrogen ion concentration and then calculate the pH from that.: [H + ] [H 2 PO 4 ] As you see from the equation, for each H there is a H2PO4. Thus you can let each = x. K a = [H 3 PO 4 ] Look up the first pK a for phosphoric acid in Appendix IV of the Segel book. (It is 2.12) and then take the antilog of 2.12 to get the Ka, which is 7.59 x 103 M. (Since the concentration of the acid is given as .2 M and it is less than 100 x the K a , you will have to include the removal of x in the denominator of the equation.) K a = 7.59 x 103 M = x 2 0.2M  x x 2 = 1.52 x 102 7.59 x 103 x x 2 + 7.59 x 103 x 1.59 x !02 = Use the quadratic equation to solve for x. x = 34.94 x 103 M = [H + ] pH =  log [H + ] =  log .03494 M = 1.457 Problem 25 (continued)...
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 Fall '08
 Hilt

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